%I #24 Aug 01 2020 11:19:12
%S 0,0,1,0,2,2,4,5,10,11,20,25,38,49,73,91,131,167,228,291,392,493,653,
%T 822,1065,1336,1714,2131,2706,3354,4209,5193,6471,7934,9817,11990,
%U 14725,17909,21875,26477,32172,38797,46893,56339,67804,81147,97260,116017
%N Number of doubletons in all partitions of n. By a doubleton in a partition we mean an occurrence of a part exactly twice (the partition [4,(3,3),2,2,2,(1,1)] has two doubletons, shown between parentheses).
%C a(n) = (the number of 2's in all partitions of n) - (the number of 3's in all partitions of n). - _Gregory L. Simay_, Jul 28 2020
%H Alois P. Heinz, <a href="/A116646/b116646.txt">Table of n, a(n) for n = 0..1000</a>
%F G.f.: x^2 / ((1+x)*(1-x^3)*(Product_{j>=1} 1-x^j)).
%F a(n) = Sum_{k>=0} k * A116644(n,k).
%F a(n) ~ exp(Pi*sqrt(2*n/3)) / (3*2^(5/2)*Pi*sqrt(n)). - _Vaclav Kotesovec_, Mar 07 2016
%e a(6) = 4 because in the partitions of 6, namely [6],[5,1],[4,2],[4,(1,1)],[(3,3)],[3,2,1],[3,1,1,1],[2,2,2],[(2,2),(1,1)],[2,1,1,1,1] and [1,1,1,1,1,1], we have a total of 4 doubletons (shown between parentheses).
%p f:= x^2/(1+x)/(1-x^3)/product(1-x^j, j=1..70): fser:= series(f,x=0,70): seq(coeff(fser,x,n), n=0..55);
%t nmax = 50; CoefficientList[Series[x^2/((1+x)*(1-x^3)) * Product[1/(1-x^k), {k, 1, nmax}], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, Mar 07 2016 *)
%t Table[Sum[PartitionsP[n-6*m-2] - PartitionsP[n-6*m-3] + PartitionsP[n-6*m-4], {m, 0, Floor[n/6]}], {n, 0, 50}] (* _Vaclav Kotesovec_, Mar 07 2016 *)
%Y Cf. A116644. Column k=2 of A197126.
%K nonn
%O 0,5
%A _Emeric Deutsch_, Feb 20 2006